Problem: Alice has a cute cat. To keep her cat fit, Alice wants to design an exercising walk for her cat! 

Initially, Alice's cat is located in a cell (x,y) of an infinite grid. According to Alice's theory, cat needs to move: 

  * exactly a steps left: from (u,v) to (u-1,v); 
  * exactly b steps right: from (u,v) to (u+1,v); 
  * exactly c steps down: from (u,v) to (u,v-1); 
  * exactly d steps up: from (u,v) to (u,v+1). 



Note that the moves can be performed in an arbitrary order. For example, if the cat has to move 1 step left, 3 steps right and 2 steps down, then the walk right, down, left, right, right, down is valid.

Alice, however, is worrying that her cat might get lost if it moves far away from her. So she hopes that her cat is always in the area [x_1,x_2]× [y_1,y_2], i.e. for every cat's position (u,v) of a walk x_1 ≤ u ≤ x_2 and y_1 ≤ v ≤ y_2 holds.

Also, note that the cat can visit the same cell multiple times.

Can you help Alice find out if there exists a walk satisfying her wishes?

Formally, the walk should contain exactly a+b+c+d unit moves (a to the left, b to the right, c to the down, d to the up). Alice can do the moves in any order. Her current position (u, v) should always satisfy the constraints: x_1 ≤ u ≤ x_2, y_1 ≤ v ≤ y_2. The staring point is (x, y).

You are required to answer t test cases independently.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^3) — the number of testcases. 

The first line of each test case contains four integers a, b, c, d (0 ≤ a,b,c,d ≤ 10^8, a+b+c+d ≥ 1).

The second line of the test case contains six integers x, y, x_1, y_1, x_2, y_2 (-10^8 ≤ x_1≤ x ≤ x_2 ≤ 10^8, -10^8 ≤ y_1 ≤ y ≤ y_2 ≤ 10^8).

Output

For each test case, output "YES" in a separate line, if there exists a walk satisfying her wishes. Otherwise, output "NO" in a separate line. 

You can print each letter in any case (upper or lower).

Example

Input


6
3 2 2 2
0 0 -2 -2 2 2
3 1 4 1
0 0 -1 -1 1 1
1 1 1 1
1 1 1 1 1 1
0 0 0 1
0 0 0 0 0 1
5 1 1 1
0 0 -100 -100 0 100
1 1 5 1
0 0 -100 -100 100 0


Output


Yes
No
No
Yes
Yes
Yes

Note

In the first test case, one valid exercising walk is $$$(0,0)→ (-1,0) → (-2,0)→ (-2,1) → (-2,2)→ (-1,2)→(0,2)→ (0,1)→ (0,0) → (-1,0)$$$

Solution (in PYTHON3):

Solution:
import sys
import os
from io import BytesIO, IOBase

#########################
# imgur.com/Pkt7iIf.png #
#########################

# returns the list of prime numbers less than or equal to n:
'''def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r'''

# returns all the divisors of a number n(takes an additional parameter start):
'''def divs(n, start=1):
    divisors = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors'''

# returns the number of factors of a given number if a primes list is given:
'''def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
    return divs_number'''

# returns the leftmost and rightmost positions of x in a given list d(if x isnot present then returns (-1,-1)):
'''def flin(d, x, default=-1):
    left = right = -1
    for i in range(len(d)):
        if d[i] == x:
            if left == -1: left = i
            right = i
    if left == -1:
        return (default, default)
    else:
        return (left, right)'''

# returns (b**p)%m 
'''def modpow(b,p,m):
    res=1
    b=b%m
    while(p):
        if p&1:
            res=(res*b)%m
        b=(b*b)%m
        p>>=1
    return res'''

# if m is a prime this can be used to determine (1//a)%m or modular multiplicative inverse of a number a
'''def mod_inv(a,m):
    return modpow(a,m-2,m)'''

# returns the ncr%m for (if m is a prime number) for very large n and r
'''def ncr(n,r,m):
    res=1
    if r==0:
        return 1
    if n-r<r:
        r=n-r
    p,k=1,1
    while(r):
        res=((res%m)*(((n%m)*mod_inv(r,m))%m))%m
        n-=1
        r-=1
    return res'''

# returns ncr%m (if m is a prime number and there should be a list fact which stores the factorial values upto n):
'''def ncrlis(n,r,m):
    return (fact[n]*(mod_inv(fact[r],m)*mod_inv(fact[n-r],m))%m)%m'''

#factorial list which stores factorial of values upto n:
'''mx_lmt=10**5+10
fact=[1 for i in range(mx_lmt)]
for i in range(1,mx_lmt):
    fact[i]=(i*fact[i-1])%mod'''

#count xor of numbers from 1 to n:
'''def xor1_n(n):
    d={0:n,1:1,2:n+1,3:0}
    return d[n&3]'''

def cel(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def prr(a, sep=' '): print(sep.join(map(str, a)))
def dd(): return defaultdict(int)
def ddl(): return defaultdict(list)
def ddd(): return defaultdict(defaultdict(int))
 

# region fastio
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
	newlines = 0
 
	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None
 
	def read(self):
			while True:
					b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
					if not b:
							break
					ptr = self.buffer.tell()
					self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
			self.newlines = 0
			return self.buffer.read()
 
	def readline(self):
			while self.newlines == 0:
					b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
					self.newlines = b.count(b"\n") + (not b)
					ptr = self.buffer.tell()
					self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
			self.newlines -= 1
			return self.buffer.readline()
 
	def flush(self):
			if self.writable:
					os.write(self._fd, self.buffer.getvalue())
					self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
	def __init__(self, file):
			self.buffer = FastIO(file)
			self.flush = self.buffer.flush
			self.writable = self.buffer.writable
			self.write = lambda s: self.buffer.write(s.encode("ascii"))
			self.read = lambda: self.buffer.read().decode("ascii")
			self.readline = lambda: self.buffer.readline().decode("ascii")
 

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

from collections import defaultdict 
#from collections import deque
#from collections import OrderedDict
#from math import gcd
#import time
#import itertools
#import timeit
#import random
#from bisect import bisect_left as bl
#from bisect import bisect_right as br
#from bisect import insort_left as il
#from bisect import insort_right as ir
#from heapq import *
#mod=998244353
#mod=10**9+7

for _ in range(ii()):
    a,b,c,d=mi()
    x,y,x1,y1,x2,y2=mi()
    if x1==x2 and (a>0 or b>0):
        print('NO')
    elif y1==y2 and(c>0 or d>0):
        print('NO')
    elif x1<=x+b-a<=x2 and y1<=y+d-c<=y2:
        print('YES')
    else:
        print('NO')